>> or(false,true)
ans =
  logical
   1

(I meant variant #5)

Ha, it looks like variant #4 got me. Quite clever. When pasted in Matlab it’s a bit more obvious because the “10< 9.9" is coded purple, indicating character arrays.

function test
    try
        if(false)
            or(true);
            disp('Yaba');
        else
            disp('Daba');
        end
    catch
        disp('Doo!');
    end
end

We never actually reach the line “or(true)”, which would throw an error, causing the catch block to print ‘Doo!’, so instead we simply print ‘Daba’. The parentheses are required in order to have the (false) alone apply to the if statement, as opposed to what comes after.

Regarding the other questions:

if (false) or (true) % variant #1
if (true) or (false) % variant #2 — my guess is this will print ‘Doo!’, since the or(false) will throw an error
if (true) or (10< 9.9) % variant #3 — will also print ‘Doo!’ for the same reason: “or(10<9.9)” is invalid
if true or 10< 9.9 % variant #4 — will also print ‘Doo!’, but this time it’s because “true or 10< 9.9” or: “10>9.9 or 10<9.9” is invalid — the or function requires 2 arguments

For backwards compatibility, space is a valid separator between the expression in the if-condition and the first statement in the body. We don’t want to encourage users to write more such code because we realize this can be very confusing, but we also don’t want to break existing code which is out there and works correctly.

Therefore, the Code Analyzer tries to provide an indication of the current behavior and direct users away from it. For this specific code example, it provides these two messages on the “or” in the MATLAB Editor:

This statement (and possibly following ones) cannot be reached.
Consider using newline, semicolon, or comma before this statement for readability.

Oh dear, seems I was wrong on (at least) two counts above…

First, the first set of brackets seem to do nothing. I was sure you normally needed a comma after the condition expression, but apparently not?!

Second, the command in variant #4 still doesn’t qualify as part of the if condition. So after having satisfied the if true condition, first the result of the unsuppressed command or 10< 9.9 is displayed in the command window:

ans =
  1×3 logical array
   1   1   1

then ‘Yaba’ is displayed!

That behaviour is of course the reason the main puzzle results in Daba, after seeing that the full condition of the if statement is just false. I don’t think that an if statement skipping the block it would execute if true and proceeding to the else block can be described as short-circuit evaluation, but maybe my understanding of execution engine theory falls short here.

The variants are fun, too.

In variant #2 we pass the if condition and get to the function call to or (disguised as a binary operator to fool users of more verbose languages than MATLAB) with only one argument – throwing an exception that results in a Doo!

In variant #3 nothing changes, because of course 10< 9.9 just evaluates to a scalar false which still isn’t an adequate input to the or function. But of course this is just a set up for

variant #4, which cunningly uses command syntax to pass two character vector arguments to or, each carefully engineered to be three characters long so that we evaluate the equivalent of ’10<‘ | ‘9.9’ which is a valid vector boolean operation. Since none of the characters involved is the null character the result is `[true true true]` so we finally get a `Yaba`.

Happy April Fools’!

new here 🙂

nice puzzle. I have never used ‘or’ like that, only used function or(A,B), or the || operator. I would have expected error there…so that was my answer 🙂
Now I’ll need to review my knowledge on syntax…

I suspect the reason for this stems from the support for a single line if. I suspect Matlab parses everything from the if to the end of the first function call (or of course || or &&) as the conditional.

%triggers the else branch:
if (true && false) or (true)
%  ^. . . . . . .^
%this is parsed as the conditional (space delimited command)
if (true && false) or (true)
%                  ^. . . .^ 
%This is considered the code to be executed on if, because there is no
%logical operator there, and the preceding part is a complete command.
%Removing the parentheses there will not make a difference, as the space is
%still demarcating the logical statement from the consequence code.

Replacing the true and if statements by anything that returns those statements does not change the parsing. Removing the parentheses will only change the or() call to be in the char mode, which would then still be missing its second input. Curiously, you can’t enter a false in char mode. If you try with eval to have a char(0) to force a false, an error is returned instead (not enough inputs), as eval seems to crop from char(0) onwards.

if (false) || (true)
etc

the parsing is:

if false,
 or(true)
 disp(1)
else
 disp(2)
end